Question

At t = 0, a series-connected capacitor and inductor are placed across the terminals of a black box as shown in (Figure 1). For t > 0, it is known that in = 1.5e-16,000t – 0.5e-4000t A, where t is in seconds. Figure < 1 of 1 25 mH + + t=0 uc 625 nF Black box

Answer 1

Solution 16ooot yoout lo o.se А. Velo) a -son lo t=0 ന്നെത 25mH + Black + ☺ 625nf box → standard relation b Current for capacitor & voltage is ۔ v Ve duc di -T but in given diagram capacitor current io e cop voltage ve not in Standard form. (current is outward from t ue terminel) duelty are t lo a C t dt velty & Solts alt Ve -

we by eq. O can find Volts across using Capacitor for tro 16000 t Velt) -0.5e {(I-se loot ose tootjalt 6254169 volt 109 1.5 16000t 0.5 _400ot. [ vocet e 7 = 16000 4000 625 +k Voll there will be a constant for îndefinate integral K is introduced whose value can be find by using initial velue V (O) 2-50v, 16ooot 4000 t tk doo Velt) 150 e Vc (0) sov but given that put tzo 150 200 tk ve (0) -sotk - 50 > .nocot - 16000 doo é 150 e Velf)

to know the voltage doop OCYOSS Now inductor for tso we will again use Standard result. ا VL + mm L Ve I di di in given pooblem VL lo mm 250H Y a Ldio It yooot ./6000t osé 2SX163 d (ise (Ise lewat e sono ✓lt) 25x103 d dt yooot 1.5x16000 é Groot to.sx 4000 astros ( (t) 16000t v - yooot 50 e 600 e + V 1+) both Velty & Yelt) for Now have found is current to due too,

+ V mom + Apply ku in the loop Vc tų tvo 2 No = Very 16000 t Vo (150 e -/Goool Gooe + 200 e soo hooof 16000t Vo yooot a (150+600) e V oot -(200 so) e 16000t Youo boot v No 750 e 250 e there otu ve a Standard result are only for Standard format of Capacitor & inductor ir pofanity Is different will be sign before formula to the sum of all vollage drop in close loop is always Zero. ** Accerding KUL, a